A first order differential equation is an equation in the form
y^\prime = f(x,y).
There is no general solution to a first order differential equation, so we’ll have to examine special cases .
Linear Equations
Consider the general first order linear equation,
y^\prime + p(x)y = g(x),
where p(x) and g(x) are continous functions on some interval. Lucky for us, a general solution exists!
The general solution for a first order linear differential equation is
y = \frac{1}{\mu(x)}\left(\int \mu(x) g(x) dx + C\right)
where
\mu(x) = \exp \left(\int p(x) dx\right)
We first need to consider a function \mu(x) such that
\mu(x) p(x) = \mu^\prime (x)
On the assumption that \mu(x) exists,
\mu(x)y^\prime + \mu^\prime(x)y = \mu(x)g(x).
Notice that the LHS of the equation looks suspiciously like the product rule, so
(\mu(x)y(x))^\prime = \mu(x)g(x).
Integrating and swapping some terms around allows us to isolate y(x)
\mu(x)y(x) + C = \int \mu(x)g(x) dx\\
y(x) = \frac{1}{\mu(x)}\left(\int \mu(x)g(x) dx + C\right)
Which gives us our general solution! We just need to find an explicit function \mu(x).
\mu(x)p(x) = \mu^\prime(x)\\
p(x) = \frac{\mu^\prime(x)}{\mu(x)}
Notice the RHS is the general solution of the derivative \ln f(x)
p(x) = \left( \ln \mu(x) \right)^\prime\\
\int p(x) = \ln \mu(x)\\
\mu(x) = \exp \left( \int p(x) dx \right)
Thus we have proved out general solution
An Initial value problem is a first order differential equation together with an initial condition, for example
\begin{cases}
y^\prime + ay = 0\\
y(0) = 2
\end{cases}
Using the formula we just proved, the solution to the first equation is simply
y = Ce^{-ax}.
But when given an initial condition, we have to pick a value of C such that y(0) = 2
y(0) = Ce^{-a0} = C \implies C = 2
So we get the answer to our IVP
y = 2e^{-ax}
Seperable Equations
Consider the equation
M(x) + N(y)y^\prime = 0
Where M(x) and N(y) are continous functions. Not every first order differential equation can be wrote like this, but if it can, we call the equation seperable.
The implicit solution for a first order seperable differential equation is given by
\int M(x) dx = \int -N(y) dy
We can seperate our equation such that all the x terms are on one side
M(x) = -N(y)y^\prime \implies M(x) = -N(y)\frac{dy}{dx}
Integrate both sides by x
\int M(x) dx = \int -N(y)\frac{dy}{dx} dx
Use a dummy substitution u = y(x), giving
du = y^\prime(x) dx = \frac{dy}{dx} dx \\
\int M(x) dx = \int -N(u) du
Then for ease of notation, simply let u = y
\int M(x) dx = \int -N(y) dy
For example
The equation
y^\prime = 6y^2x
can be seperated into
\frac{y^\prime}{y^2} - 6x = 0.
Thus, M(x) = 6x and N(y) = \frac{1}{y^2}, and we get our implicit solution
\int -6x dx = \int -\frac{1}{y^2} dy\\
-3x^2 + C = \frac{1}{y}\\
Which can easily be converted into our explicit solution
y = \frac{1}{C - 3x^2}